LDR Based Input Selection

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A few years ago Jon VerHalen of Lowther America bought some LDRs from me and asked if I had ever thought to use them for input selection. He sent me a simple schematic with a series LDR and a relay after it. The idea is to let the LDR go dark, 25Megohms, and shunt anything that gets through to ground. I added a series LDR after the relay because if I am sending signal from another source I dont want it to get shunted to ground by the unselected source's relay. That and a good power supply is about all you need. So it goes :
Series LDR/shunt to ground/Series LDR
25Megohm/5ohm/25Megohm for non-selection of a source
40Ohm/infinite Ohm/40Ohm for selection of a source
Yes there is series resistance but it just ends up being part of the series resistance of your attenuator anyway and 80R is nothing to a 10k pot. If you turn your attenuator up all the way you are only missing the last 1% of volume but you have an incredibly great sounding input selector.
 
Some pics in various stages of completion
 

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Here is the schematic. Simple as it may be.
This is the general idea. Regulated voltage is then put through a current source which supplies power to the LDRs. Regulated voltage is applied to a trimmer that turns on the transistors. By twiddling the trimmer you can balance the resistance of the LDRs so they are the same. By increasing the voltage you can lower the resistance of both at the same time.
ON means both LDRs turn on full force at 40R each for a series total of 80R and the relay is open.
OFF means that both LDRs have 25Megohm or more resistance and the relay has less than 5R. Its an optocoupler relay to keep with the theme.
 

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A pair of LDRs makes for a great switch. I think the traditional "switch" application for LDRs is just a simple shunt where both elements are LDRs. Either one is turned on or the other. This gives two states: very high input impedance (like 25M ohms) and a very low impedance between that and ground (e.g. the OFF state) or very low input impedance and a very high impedance between that and ground (e.g. the ON state).

Using LDRs as 100% on/off devices results in very low distortion IIRC according to the application note. Much less than when used as an attenuator.

I can't follow why you used this particular circuit (in post #6) for switching...

-Charlie
 
Charlie
If we have 4 inputs all going to the same attenuator and we have a series element and shunt element and not a second series element....
If we have, for example, source 1 ON and 2, 3 and 4 OFF then source one has 40R series followed by infinte ohms to ground and then it touches the series resistor of the attenuator.
If 2, 3 and 4 are OFF then they would have their shunt element with only minimal resistance to ground and still their output is touching the series resistor of the attenuator.
If we send signal though Source 1 at this point it will all go direct to ground through the shunt elements of 2, 3 and 4 without the series LDR after the shunt element.
Its not a rotary switch or a relay. They are all connected all the time so we have to isolate the grounds when we want signal to pass.
 
Charlie
If we have 4 inputs all going to the same attenuator and we have a series element and shunt element and not a second series element....
If we have, for example, source 1 ON and 2, 3 and 4 OFF then source one has 40R series followed by infinte ohms to ground and then it touches the series resistor of the attenuator.
If 2, 3 and 4 are OFF then they would have their shunt element with only minimal resistance to ground and still their output is touching the series resistor of the attenuator.
If we send signal though Source 1 at this point it will all go direct to ground through the shunt elements of 2, 3 and 4 without the series LDR after the shunt element.
Its not a rotary switch or a relay. They are all connected all the time so we have to isolate the grounds when we want signal to pass.

OK, what about connecting the inputs in between two LDRs (like a voltage divider) that connect to the attenuator at one end and ground at the other. This would provide the necessary isolation of non-selected signals by 25M ohms, although the input signals would be grounded.

Anyway, I guess what you are doing is fine. If you didn't care that the input signal was grounded when it was not selected, you could do it with just 2 LDRs and get rid of the relays. What's your take on that?

-Charlie
 
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Charlie
A few years back I was listening to music and had been infatuated with how different impedances of passive pots sounded different. I was using my Lighter Note LDR attenuator and at the moment was crazy about the sound. I wanted to know what is the impedance RIGHT NOW. So without thinking, without turning it off or unhooking anything I put my DMM across the series LDR's resistor. Realized instantly that there was DC from my DMM going to the amp and removed the DMM. In that moment right before removing it something sounded better. What was it? I thought for a day or so and in the shower the next morning decided that only high resistance and a low amount of DC came from the DMM so I measured my DMM with another and found 33M resistance.
I bought many of these resistors from Mouser. 249k-33M and put them in parallel with the series LDR. All of them individually made a difference and many of them together made a difference. More air, transparency, highs were better. My guess is it was bypassing a filter created by the capacitance in the cables and the resistance of the pot. Thats the long of it. The short of it is that even through 25M you can have music if there is no short to ground after that 25M.
Your idea would work as well but then we would have signal coming in and choosing between 25M and 40R. Some will still choose 25M. It would not work like a voltage divider because in your schematic the 40R to ground would still come before the 25M.
I think I understood your point. Correct me if I dont.
Now, if we were to not ground but just use 25M then my experiment with the 33M resistor comes into play. You would still hear the other sources. Example, you can turn the Lighter Note or a Lightspeed off and all LDRs go dark but you can still hear the music. Awful distorted music.
Uriah
 
Here's what I am thinking of:

Take the figure below, and connect it "backwards":
An externally hosted image should be here but it was not working when we last tested it.


My idea is to connect the inputs to the "Vout" terminal in the figure, and have the "Vin" terminal going to your attenuator.

Use two LDRs for the resistive elements, and switch one or the other "on" while the remaining one is "off". This would produce the following:
  • The input impedance of this stage would be either 40R (when not selected) or 25M (when selected)
  • Unselected signals are essentially grounded by the ratio of 40R/(40R+25M), resulting in an attenuation of 116dB. Also, there is a 25M+40R ohm impedance in between the attenuator and ground for each signal, so for a modest number of inputs there is little loss.
  • Selected signals are attenuated only by 40R/(40R+25M), which works out to only 0.000014dB, e.g. basically no attenuation.

This is different from putting a 25M LDR resistance in series only. In that case there would be some current that still makes it through.

-Charlie
 
Charlie,
Actually that would not work as well as you or I were thinking.
If my thinking is right, the problem is that the resistor going to ground, if we reverse that schematic, is now before the series resistor. No longer a voltage divider. Voltage division would happen between the output resistor of the source and the shunt to ground and the resulting voltage would travel through the series LDR.
Uriah

edit:I just reproduced your reverse attenuator circuit with a 56R and 1M. Put 2V through it and got 1.8V out.
Then I did 20meg series/56Rshunt/20meg series and my 6position DMM cant measure it.
 
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Charlie,
Actually that would not work as well as you or I were thinking.
If my thinking is right, the problem is that the resistor going to ground, if we reverse that schematic, is now before the series resistor. No longer a voltage divider. Voltage division would happen between the output resistor of the source and the shunt to ground and the resulting voltage would travel through the series LDR.
Uriah

OK, not exactly like a traditional voltage divider on the input, that's true. But it should function as a switch like I described. What you are really concerned about is the level of each input as seen from the attenuator (volume control) further downstream. When the input is not selected, the input signal is essentially connected to ground via 40R while a 25M series resistance is put in between the input and the volume control. The current from the input will flow to ground and no current will flow to the volume control, e.g. there will be 116dB rejection. The series LDR provides switchable isolation between inputs and volume control, and the grounded LDR is directing the input current.

-Charlie
 
The current is important Charlie but the voltage is what carries the information and is multiplied by the gain of the amp. Your solution will not function as a switch. It allows literally millions of times more voltage through. It doesnt divide voltage and thats what is necessary. I just ran the tests on the bench. Nearly no attenuation with shunt first.
 
edit:I just reproduced your reverse attenuator circuit with a 56R and 1M. Put 2V through it and got 1.8V out.
Then I did 20meg series/56Rshunt/20meg series and my 6position DMM cant measure it.

Right. I think that the DMM does not present enough of a load. All you are measuring is the loading down of your 2V supply through the 56R.

Can you try adding a 10k or 20k resistor as shown below:

input---LDR1---ground
input---LDR2---(measurement point)---20k---ground

Connect the inputs together and also connect the 2V source there.

When LDR1 = 25M and LDR2 = 50R the switch will be "on" and you should measure 2V.
When LDR1 = 50R and LDR2 = 25M the switch is "off" and you should measure near 0V

The 20k resistor represents the total resistance of the voltage divider volume control.

If that doesn't work, your idea sure seems to. I think that the input LDR and relay are essentially making a voltage divider with the relay switching in a short or a large resistance.

-Charlie
 
The current is important Charlie but the voltage is what carries the information and is multiplied by the gain of the amp. Your solution will not function as a switch. It allows literally millions of times more voltage through. It doesnt divide voltage and thats what is necessary. I just ran the tests on the bench. Nearly no attenuation with shunt first.

no current = no signal.
 
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